Two capacitors of 0.0003μF and 0.0006μF are connected in series, find their combined capacitance.

When capacitors are connected in series, the formula to find their combined capacitance \(C_{\text{total}}\) is given by:

\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \]

where \(C_1\) and \(C_2\) are the capacitances of the individual capacitors. In this case, \(C_1 = 0.0003 \, \mu\text{F}\) and \(C_2 = 0.0006 \, \mu\text{F}\).

First, calculate the reciprocal of each capacitance:

\[ \frac{1}{C_1} = \frac{1}{0.0003} \]

\[ \frac{1}{C_2} = \frac{1}{0.0006} \]

Calculating each value:

\[ \frac{1}{0.0003} = \frac{10^6}{3} \] and \[ \frac{1}{0.0006} = \frac{10^6}{6} \]

Now, add these values together:

\[ \frac{1}{C_{\text{total}}} = \frac{10^6}{3} + \frac{10^6}{6} = \frac{10^6 \times 2}{6} + \frac{10^6 \times 1}{6} = \frac{10^6 \times 3}{6} = \frac{10^6}{2} \]

Finally, take the reciprocal of the resulting value to find \(C_{\text{total}}\):

\[ C_{\text{total}} = \frac{2}{10^6} = 0.0002 \, \mu\text{F} \]

So, the combined capacitance of the two capacitors in series is 0.0002 μF.